Equilibrium Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A weak acid \( \ce{HX} \) (\( K_a = 3.2 \times 10^{-5} \)) is mixed with 0.04 M \( \ce{NaX} \) in a 2:1 volume ratio (acid:salt). If \( [\ce{HX}] = 0.06 \, \text{M} \) after mixing, what is the pH?

Total volume = 3V, \( [\ce{X-}] = \frac{0.04V}{3V} = 0.0133 \, \text{M} \), \( \text{p}K_a = 4.5 \), \( \text{pH} = 4.5 + \log \frac{0.0133}{0.06} = 4.5 - 0.655 = 4.85 \).

4.5
5.2
4.85
4.0
3

What is the pH of a solution containing 0.01 M \( \ce{HCl} \) assuming complete dissociation?

For \( \ce{HCl} \), \( [\ce{H+}] = 0.01 \, \text{M} = 10^{-2} \, \text{M} \), so \( \text{pH} = -\log(10^{-2}) = 2 \).

1
2
3
4
2

In a saturated solution of a sparingly soluble salt \( \ce{MX2} \), the solubility is \( S \) mol/L. If \( K_{sp} = 4S^3 \), what is the relationship between \( \ce{M^{2+}} \) and \( \ce{X^-} \)?

For \( \ce{MX2 <=> M^{2+} + 2X^-} \), \( [\ce{M^{2+}}] = S \), \( [\ce{X^-}] = 2S \), so \( K_{sp} = S (2S)^2 = 4S^3 \). Thus, \( [\ce{X^-}] = 2[\ce{M^{2+}}] \).

\( [\ce{X^-}] = [\ce{M^{2+}}] \)
\( [\ce{X^-}] = 2[\ce{M^{2+}}] \)
\( [\ce{M^{2+}}] = 2[\ce{X^-}] \)
\( [\ce{X^-}] = 4[\ce{M^{2+}}] \)
2

For the reaction \( \ce{2A(g) + B(g) <=> 3C(g)} \), \( K_c = 64 \) at 300 K. If 2 moles of \( \ce{A} \) and 1 mole of \( \ce{B} \) are placed in a 2 L vessel, what is \( [\ce{C}] \) at equilibrium?

Initial: \( [\ce{A}] = \frac{2}{2} = 1 \, \text{M} \), \( [\ce{B}] = \frac{1}{2} = 0.5 \, \text{M} \), \( [\ce{C}] = 0 \). Let \( 3x \) be moles of \( \ce{C} \) formed, so \( \ce{A} \) decreases by \( 2x \), \( \ce{B} \) by \( x \). At equilibrium: \( [\ce{A}] = 1 - x \), \( [\ce{B}] = 0.5 - x \), \( [\ce{C}] = 1.5x \). \( K_c = \frac{[\ce{C}]^3}{[\ce{A}]^2[\ce{B}]} = \frac{(1.5x)^3}{(1 - x)^2(0.5 - x)} = 64 \). Simplifying, \( \frac{3.375x^3}{(1 - x)^2(0.5 - x)} = 64 \). Trial: \( x = 0.4 \), \( [\ce{A}] = 0.6 \), \( [\ce{B}] = 0.1 \), \( [\ce{C}] = 0.6 \), \( \frac{(0.6)^3}{(0.6)^2(0.1)} = \frac{0.216}{0.036} = 6 \), too small. Solving approximately, \( x \approx 0.48 \), \( [\ce{C}] = 1.5 \times 0.48 = 0.72 \, \text{M} \).

0.72 M
0.96 M
0.48 M
1.44 M
1

A weak base \( \ce{BOH} \) (\( K_b = 5.0 \times 10^{-6} \)) has a 0.1 M solution with \( \alpha = 0.00707 \). What is the pH?

\( [\ce{OH-}] = 0.1 \times 0.00707 = 7.07 \times 10^{-4} \), \( \text{pOH} = 3.15 \), \( \text{pH} = 14 - 3.15 = 10.85 \).

3.15
11.0
10.0
10.85
4

For \( \ce{2A(g) <=> B(g) + C(g)} \), \( K_c = 0.0625 \) at 300 K. If 0.8 mol \( \ce{A} \) is placed in a 2 L vessel with 0.1 mol \( \ce{B} \), what is \( [\ce{C}] \) at equilibrium?

Initial: \( [\ce{A}] = \frac{0.8}{2} = 0.4 \, \text{M} \), \( [\ce{B}] = \frac{0.1}{2} = 0.05 \, \text{M} \), \( [\ce{C}] = 0 \). Let \( x = [\ce{C}] \), \( [\ce{A}] = 0.4 - 2x \), \( [\ce{B}] = 0.05 + x \). \( K_c = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]^2} = \frac{(0.05 + x)x}{(0.4 - 2x)^2} = 0.0625 \). Solving, \( (0.05 + x)x = 0.0625 (0.4 - 2x)^2 \), test \( x = 0.05 \): \( (0.05 + 0.05) \times 0.05 = 0.005 \), \( 0.0625 \times (0.4 - 0.1)^2 = 0.005625 \) (close), \( x \approx 0.05 \, \text{M} \).

0.05 M
0.1 M
0.025 M
0.2 M
1

For the reaction \( \ce{2A(g) + 2B(g) <=> 3C(g)} \), \( K_c = 64 \) at 500 K. If 2 moles of \( \ce{A} \) and 2 moles of \( \ce{B} \) are placed in a 1 L vessel, what is \( [\ce{C}] \) at equilibrium?

Initial: \( [\ce{A}] = 2 \, \text{M} \), \( [\ce{B}] = 2 \, \text{M} \), \( [\ce{C}] = 0 \). Let \( 3x \) be moles of \( \ce{C} \) formed, so \( \ce{A} \) and \( \ce{B} \) decrease by \( 2x \). At equilibrium: \( [\ce{A}] = 2 - 2x \), \( [\ce{B}] = 2 - 2x \), \( [\ce{C}] = 3x \). \( K_c = \frac{[\ce{C}]^3}{[\ce{A}]^2[\ce{B}]^2} = \frac{(3x)^3}{(2 - 2x)^2 (2 - 2x)^2} = \frac{27x^3}{(2 - 2x)^4} = 64 \), \( \frac{27x^3}{(2 - 2x)^4} = 64 \), \( \frac{3x}{2 - 2x} = 4 \), \( 3x = 8 - 8x \), \( 11x = 8 \), \( x \approx 0.727 \), \( [\ce{C}] = 3 \times 0.727 \approx 2.18 \, \text{M} \).

2.18 M
1.5 M
3.0 M
0.727 M
1

The \( K_a \) of \( \ce{CH3COOH} \) is \( 1.8 \times 10^{-5} \). What is the \( K_b \) of \( \ce{CH3COO-} \) at 298 K?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), \( K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \).

\( 1.8 \times 10^{-5} \)
\( 1.0 \times 10^{-14} \)
\( 9.0 \times 10^{-10} \)
\( 5.56 \times 10^{-10} \)
4

For \( \ce{A2(g) <=> 2A(g)} \), \( K_c = 0.09 \) at 300 K. If 0.2 mol \( \ce{A2} \) is in a 1 L vessel, what is the degree of dissociation?

Initial: \( [\ce{A2}] = 0.2 \, \text{M} \). Let \( \alpha \) be the degree of dissociation, \( [\ce{A2}] = 0.2 (1 - \alpha) \), \( [\ce{A}] = 0.4\alpha \). \( K_c = \frac{[\ce{A}]^2}{[\ce{A2}]} = \frac{(0.4\alpha)^2}{0.2 (1 - \alpha)} = 0.09 \), \( 0.8\alpha^2 = 0.09 (1 - \alpha) \), \( \alpha \approx 0.3 \).

0.1
0.5
0.2
0.3
4

Which of the following is a Lewis base?

A Lewis base donates an electron pair. \( \ce{CN-} \) has a lone pair on carbon or nitrogen, making it a Lewis base.

\( \ce{BCl3} \)
\( \ce{AlCl3} \)
\( \ce{CN-} \)
\( \ce{H+} \)
3

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